Please show all work in order to receive 5 stars. Thanks!! limx rightarrow 0(1+x

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Explanation / Answer

Take the limit: lim_(x-%u026C) (-1/(-1+e^x)+1/x) Indeterminate form of type infinity - infinity, write -1/(-1+e^x)+1/x as -(e^x (e^x-(e^x-x)^2-x))/((e^x-e^(2 x)) x (-e^x+x)): = lim_(x->0) -(e^x (e^x-(e^x-x)^2-x))/((e^x-e^(2 x)) x (-e^x+x)) Factor out constants: = -(lim_(x->0) (e^x (e^x-(e^x-x)^2-x))/((e^x-e^(2 x)) x (-e^x+x))) Factor the numerator and denominator: = -(lim_(x->0) ((e^x (-e^x+x)) (-1+e^x-x))/((e^x (-e^x+x)) (-(-1+e^x) x))) Cancel terms: = -(lim_(x->0) -(-1+e^x-x)/((-1+e^x) x)) Factor out constants: = lim_(x->0) (-1+e^x-x)/((-1+e^x) x) Indeterminate form of type 0/0. Applying L'Hospital's rule we have, lim_(x->0) (-1+e^x-x)/((-1+e^x) x) = lim_(x->0) (( d(-1+e^x-x))/( dx))/(( d((-1+e^x) x))/( dx)): = lim_(x->0) (-1+e^x)/(-1+e^x (1+x)) Indeterminate form of type 0/0. Applying L'Hospital's rule we have, lim_(x->0) (-1+e^x)/(-1+e^x (1+x)) = lim_(x->0) (( d(-1+e^x))/( dx))/(( d(-1+e^x (1+x)))/( dx)): = lim_(x->0) 1/(2+x) The limit of a quotient is the quotient of the limits: The limit of a constant is the constant: = 1/(lim_(x->0) (2+x)) The limit of 2+x as x approaches 0 is 2: Answer: | | = 1/2

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